Problem: $f(x) = x+3$ $h(x) = 6x^{2}+x+2-2(f(x))$ $ f(h(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = 6(0^{2})+2-2(f(0))$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = 3$ $f(0) = 3$ That means $h(0) = 6(0^{2})+2+(-2)(3)$ $h(0) = -4$ Now we know that $h(0) = -4$ . Let's solve for $f(h(0))$ , which is $f(-4)$ $f(-4) = -4+3$ $f(-4) = -1$